Separable equations

First-order differential equations
The general first-order differential equation is written as
dy/dx=F(x,y) or M(x,y)dx+N(x,y)dy=0
1. Separable equations
A first-order differential equation is separable if it can be written in the form
f(x)dx=g(y)dy
where the function f(x) is independent of y and g(y) is independent of x.
We can find the general solution of this differential equation by integral

??f(x)dx=??g(y)dy
Example 1: Solve the differential equation xdy- 3ydx=0.
Solution:
Rearranging, we have
( dy )/y=3dx/x
??( dy )/y=3??dx/x
ln?y=3 ln?x+ln?C
ln?y=ln??x^3 ?+ln?C
ln?y=ln??Cx^3 ?
y=Cx^3






Example 2: Find the general solution of the differential equation y^ +2e^x y=e^x y^2.
Solution : From the given equation, we have y^ =(y^2-2y) e^x
e^x dx=1/y(y-2) dy
1/y(y-2) =( A )/y+B/(y-2)
y=0 ? A=-( 1 )/( 2 ) , y=2 ? B=( 1 )/( 2 )
??? e^x dx?=??[1/2(y-2) -( 1 )/2y]dy
e^x+C_1=( 1 )/( 2 ) ln?(y-2)-( 1 )/( 2 ) ln?y
2e^x+2C_1=ln?(y-2)-ln?y
2e^x+C=ln?((y-2)/y)


Example 3: Solve with y(1)=6
Solution: We change to differential form, separate the variables, and integrate
dy/dx=( 5 )/( 2 )-( 1 )/( 2 ) y ? dy/dx=-( y-5 )/( 2 )
?_6^y?dy/(y-5)=-( 1 )/( 2 ) ?_1^x??dx ?
? / ln?(y-5) ?|_6^y=-? ( 1 )/( 2 ) x?|_1^x
ln?(y-5)-ln?1=-( 1 )/( 2 ) x+( 1 )/( 2 ) ? ln?(y-5)=( 1-x )/( 2 )
y-5=e^(((1-x))?2) ? y=5+e^(((1-x))?2)


Example 4: Solve e^x tan?y dx=(1+e^x ) sec^2?y dy with y(ln?2 )=??4
Solution:
?_(??4)^y?tan?y/sec^2?y dy=?_ln?2^x?e^x/(1+e^x ) dx
? ? / ln?(tan?y ) ?|_(??4)^y=ln?(1+e^x ) ?|_ln?2^x
ln?(tan?y )-ln?(1)=ln?(1+e^x )-ln?(1+e^ln?2 )
ln?(tan?y )=ln?(1+e^x )-ln?(3)
ln?(tan?y )=ln??(1+e^x)/3?
tan?y=(1+e^x)/3
y=tan^(-1)?((1+e^x)/3)


Solve the ODEs:
(1) ? y?^ +2x?(?1-y?^2 )=0
(2) x^2 dy+xydx=(x+6)dy+2ydx
(3) (1+x^2 )dy-(1+y^2 )dx=0 with y(0)=1
(4) dy/dx=(y^2-y-2)/(x^2+x) with y(1)=3